AF+BG theorem and Cohen Macaulay property I tried to solve the following exercise in Vakil's notes

Question 1: According to the hint, I should try to show the intersection of affine cones is CM (since one dimensional scheme is CM iff it has no embedded points). This seems very plausible to me for it should be some affine lines glued together at the origin, how can I make this argument precise?
Question 2: Then what? If I have another homogeneous polynomial $H$, I hope to show that it is $0$ in $\mathbb{C}[x_0,x_1,x_2]/(f,g)$. How does CM (or no embedded point) help me win?
 A: The assumption is that $V_+(f,g)\subset\mathbb{P}_{\mathbb{C}}^2$, as a scheme, is 0-dimensional and reduced.
In general, if an ideal $I\subset S:=\mathbb{C}[x_0,\ldots,x_n]$ defines a reduced subscheme $V_+(I)\subset\mathbb{P}_{\mathbb{C}}^n$, it is not necessarily true that $I$ is radical. The ideal which is radical is the saturation $I':=\{f\in k[x_0,\ldots,x_n]: fS_k\subset I,\ \mbox{for some } k\}$. $I'$ is the largest ideal defining the same subscheme of $\mathbb{P}_{\mathbb{C}}^n$ as $I$.
Returning to the exercise (now $S=\mathbb{C}[x_0,x_1,x_2]$) , if $h$ is zero along $V_+(f,g)$, then $h$ is in the saturation of $(f,g)$, that is, $h S_k\subset (f,g)$ for some $k$. If $h\notin(f,g)$, then $k\geq 1$ and the latter condition implies clearly that $\mathfrak{m}=(x_0,x_1,x_2)$ is an associated prime of $S/(f,g)$.
However, $f$ is irreducible and the class of $g$ in the domain $S/(f)$ is non-zero. Therefore, $(f,g)$ is a regular sequence and hence $S/(f,g)$ is a 1-dimensional CM ring; in particular, $\mathfrak{m}$ cannot be an associated prime.
